The set of real values of x for which 2 l o g √ 2 ( x − 1 ) > x + 5 is Options: A . &nbsp ( − ∞ , − 1 ) ∪ ( 4 , + ∞ ) B . &nbsp ( 4 , + ∞ ) C . &nbsp ( − 1 , 4 ) D . &nbsp N o n e o f t h e s e

Answer: Option B : B 2 l o g √ 2 ( x − 1 ) > x + 5 ⇒ 2 l o g 2 ( x − 1 ) 2 > x + 5 ⇒ ( x − 1 ) 2 > x + 5 ⇒ x 2 − 3 x − 4 > 0 ⇒ ( x − 4 ) ( x + 1 ) > 0 ⇒ x > 4 o r x < − 1 But for given log to be defined, x - 1 > 0 i . e . , x > 1 ∴ x > 4 ⇒ x ϵ ( 4 , ∞ ) .

The set of real values of x for which 2log√2 (x−1)>x+5 i

Discussion Forum : Inequalities Modulus And Logarithms

Question -

The set of real values of x for which $2^{{l o g}_{\sqrt{2}} (x - 1)} > x + 5$ is

Options:

A .

(- \infty, - 1) \cup (4, + \infty)

B .

(4, + \infty)

C .

(- 1, 4)

D .

N o n e o f t h e s e

Answer: Option B
:
B

2^{{l o g}_{\sqrt{2}} (x - 1)} > x + 5 \Rightarrow 2^{l o g_{2} (x - 1)^{2}} > x + 5 \Rightarrow (x - 1)^{2} > x + 5 \Rightarrow x^{2} - 3 x - 4 > 0 \Rightarrow (x - 4) (x + 1) > 0 \Rightarrow x > 4 o r x < - 1

But for given log to be defined, x - 1 > 0

i . e ., x > 1 ∴ x > 4 \Rightarrow x ϵ (4, \infty) .

Was this answer helpful ?

Next Question

Discussion Forum : Inequalities Modulus And Logarithms

Submit Your Solution hear: